Question

A single phase step-down transformer has a high voltage primary winding and a low voltage secondary winding. The secondary winding is connected to a load impedance \(Z_2\). The impedance \(Z_2\) when referred to the primary winding is denoted by \(Z_2'\). Which of the following conditions is true?

• A. \(Z_2' = Z_2\)
• B. \(0.5\,Z_2 < Z_2' < Z_2\)
• C. \(Z_2' > Z_2\)
• D. \(Z_2' \leq 0.5\,Z_2\)

Hint:

Recall that an impedance on the secondary is referred to the primary by multiplying by the square of the turns ratio, \(Z_2' = a^2 Z_2\) with \(a = N_1/N_2\).

Answer 1

The Correct Option is C

Step 1: Understanding the Question.
We have a step-down transformer, so the primary (high voltage) side has more turns than the secondary (low voltage) side. A load impedance \(Z_2\) sits on the secondary. We need to know how \(Z_2\), when moved (referred) to the primary side and called \(Z_2'\), compares in size to the original \(Z_2\).

Step 2: Key Formula or Approach.
For an ideal transformer with turns ratio \(a = \frac{N_1}{N_2}\) (primary turns over secondary turns), any impedance on the secondary appears on the primary side scaled by \(a^2\):
\[ Z_2' = a^2 \, Z_2 \]
This comes from the fact that voltage scales by \(a\) and current scales by \(\frac{1}{a}\) when moving from secondary to primary, so impedance (voltage divided by current) scales by \(a \times a = a^2\).

Step 3: Detailed Explanation.
Since this is a step-down transformer, the primary winding is the high voltage, high turns side and the secondary winding is the low voltage, low turns side. That means \(N_1 > N_2\), so the turns ratio \(a = \frac{N_1}{N_2} > 1\).
Because \(a > 1\), squaring it keeps it greater than 1 as well, so \(a^2 > 1\).
Applying this to the referral formula, \(Z_2' = a^2 Z_2\) with \(a^2 > 1\) directly gives \(Z_2' > Z_2\).

Step 4: Final Answer.
Option (A) would only hold if \(a = 1\), which is not the case for a step-down transformer since the turns ratio is not unity. Options (B) and (D) both require \(Z_2' < Z_2\) (a fraction of \(Z_2\)), which happens only when referring an impedance backward for a step-up transformer, not here. Since \(a^2 > 1\) for this step-down transformer, the impedance referred to the primary is always larger than the impedance on the secondary.
\[ \boxed{Z_2' > Z_2} \]